Question Statement
Evaluate the integral:
β«(xβ1)(xβ2)(xβ3)3x2β12x+11β,dx
Background and Explanation
This problem involves the method of partial fractions, which is used to decompose a rational function into simpler fractions that can be integrated easily. Here, the denominator is already factored, and the numerator is of lower degree than the denominator, making the function suitable for partial fraction decomposition.
Solution
We aim to decompose the given fraction:
(xβ1)(xβ2)(xβ3)3x2β12x+11β.
We express the fraction as:
(xβ1)(xβ2)(xβ3)3x2β12x+11β=xβ1Aβ+xβ2Bβ+xβ3Cβ.
Step 2: Multiply to Eliminate Denominators
Multiply through by the denominator (xβ1)(xβ2)(xβ3):
3x2β12x+11=A(xβ2)(xβ3)+B(xβ1)(xβ3)+C(xβ1)(xβ2).
Step 3: Solve for Constants A, B, and C
- Let x=1:
3(1)2β12(1)+11=A(1β2)(1β3),
3β12+11=A(β1)(β2),
2=2AβΉA=1.
- Let x=2:
3(2)2β12(2)+11=B(2β1)(2β3),
3(4)β24+11=B(1)(β1),
12β24+11=βBβΉβ1=βBβΉB=1.
- Let x=3:
3(3)2β12(3)+11=C(3β1)(3β2),
3(9)β36+11=C(2)(1),
27β36+11=2CβΉ2=2CβΉC=1.
Step 4: Rewrite the Integral
Substitute A, B, and C into the partial fraction decomposition:
(xβ1)(xβ2)(xβ3)3x2β12x+11β=xβ11β+xβ21β+xβ31β.
The integral becomes:
β«(xβ1)(xβ2)(xβ3)3x2β12x+11β,dx=β«xβ11β,dx+β«xβ21β,dx+β«xβ31β,dx.
Step 5: Evaluate Each Term
- For β«xβ11β,dx:
lnβ£xβ1β£.
- For β«xβ21β,dx:
lnβ£xβ2β£.
- For β«xβ31β,dx:
lnβ£xβ3β£.
Combine results:
β«(xβ1)(xβ2)(xβ3)3x2β12x+11β,dx=lnβ£xβ1β£+lnβ£xβ2β£+lnβ£xβ3β£+C,
where C is the constant of integration.
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Partial Fraction Decomposition:
Q(x)P(x)β=xβ1Aβ+xβ2Bβ+xβ3Cβ,
for distinct linear factors in the denominator.
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Basic Logarithmic Integrals:
β«xβa1β,dx=lnβ£xβaβ£+C.
Summary of Steps
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Decompose the fraction into partial fractions:
(xβ1)(xβ2)(xβ3)3x2β12x+11β=xβ11β+xβ21β+xβ31β.
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Integrate each term individually:
β«xβ11β,dx,β«xβ21β,dx,β«xβ31β,dx.
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Combine results to obtain the final solution:
lnβ£xβ1β£+lnβ£xβ2β£+lnβ£xβ3β£+C.